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If $f(x)=\tan ^{-1}\left(\frac{\sin x}{1+\cos x}\right)$, then $f^{\prime}\left(\frac{\pi}{3}\right)=$
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$\frac{1}{2}$
$\begin{aligned} & f(x)=\tan ^{-1}\left(\frac{\sin x}{1+\cos x}\right)=\tan ^{-1}\left[\tan \frac{x}{2}\right]=\frac{x}{2} \\ & \Rightarrow f(x)=\frac{1}{2} . \text { Hence } f^{\prime}\left(\frac{\pi}{3}\right)=\frac{1}{2} .\end{aligned}$
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